solution for question 8

7 x 7 + 2 x 2 = 53, 5 x 5 + 3 x 3 = 34, 3 x 3 + 4 x 4 = 25.

The next tow numbers are 2 x 2 + 5 x 5 = 29, 2 x 2 + 9 x 9 = 85

Question 8

Number 53 is derived from 72. 34 is derived from 53. 25 is derived from 34. What are the next 2 numbers in the sequence?

solution for question 7

Perimeter = Distance covered in 8 min. =		12000	x 8	m = 1600 m.
		60

Let length = 3x metres and breadth = 2x metres.

Then, 2(3x + 2x) = 1600 or x = 160.

 Length = 480 m and Breadth = 320 m.

 Area = (480 x 320) m² = 153600 m².

Question 7

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:


answer the questions through comment

solution for question 6

Let C = x.

Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.

So, x + x + 5000 + x + 9000 = 50000

 3x = 36000

 x = 12000

A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.

A's share = Rs.		35000 x	21		= Rs. 14,700.
			50

Question 6

A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:

solution for question 5

Let C.P. be Rs. x.

Then,	1920 - x	x 100 =	x - 1280	x 100
	x		x

 1920 - x = x - 1280

 2x = 3200

 x = 1600

Required S.P. = 125% of Rs. 1600 = Rs.		125	x 1600		= Rs 2000.
		100

Question 5

The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?

solution for question 4

Let the man's rate upstream be x kmph and that downstream be y kmph.

Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.

		x x 8	4		= (y x 4)
			5

	44	x =4y
	5

y =	11	x.
	5

Required ratio =		y + x		:		y - x
		2				2

=		16x	x	1		:		6x	x	1
		5		2				5		2

=	8	:	3
	5		5

   = 8 : 3

Question 4

A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

solution for question 3

Let the speed of the stream be x km/hr. Then,

Speed downstream = (15 + x) km/hr,

Speed upstream = (15 - x) km/hr.

	30	+	30	= 4	1
	(15 + x)		(15 - x)		2

	900	=	9
	225 - x2		2

 9x² = 225

 x² = 25

 x = 5 km/hr.

Question 3

A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. Whats the speed of the stream (in km/hr)?

solution for question 2

Speed downstream = (13 + 4) km/hr = 17 km/hr.

Time taken to travel 68 km downstream =		68	hrs = 4 hrs.
		17

So, Ans is 4Hrs.

Question 2

A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

solution for question 1

On the first day, the snail reaches a height of 5 meters and slides down 4 meters at night, and thus ends at a height of 1 meter. On the second day, he reaches 6 m., but slides back to 2 m. On the third day, he reaches 7 m., and slides back to 3 m. ... On the fifteenth day, he reaches 19 m., and slides back to 15 m. On the sixteenth day, he reaches 20 m., so now he is at the top of the pit! Conclusion: The snail reaches the top of the pit on the 16th day!... . 

Question 1

A snail is at the bottom of a 20 meters deep pit. Every day the snail climbs 5 meters upwards, but at night it slides 4 meters back downwards. How many days does it take before the snail reaches the top of the pit?

About this blog

My blog was started with an intention to tingle your IQ and bring out your intelligence to the fullest.
I will be asking you mind boggling questions everyday in this page. now all you gotta do is find out the answers and post it on this blog with your name and your place through comment. The questions will be asked at 7.00PM and the answers will be published at 11.00PM everyday.  I will be announcing the winners of each day.
The blog will active from 10/10/2010 (Sun)


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